In the Michelson and Morley experiment, we predict with Galilean relativity and the assumption of the existence of a luminiferous aether that there should be a time difference between the two beams of light reaching the detector of an interferometer with arms of equal length. This time difference is given by:

$$ \Delta t = \frac{2L}{c} (\gamma^2 - \gamma) $$

where $\gamma$ is the Lorentz constant, $c$ is the speed of light from the IRF of the aether, and $L$ is the length of the length of two arms of the interferometer.

This was derived under the assumption that the IRF of the experiment ("Earth") moves through the Aether at a velocity $V$ in a single direction as shown in the diagram below.

enter image description here

The derivation was similar to the derivation on Wikipedia page here which also uses this assumption. My question is, why is this assumption is allowed when surely the Earth could be moving with a velocity in the y-direction as well?


1 Answer 1


You can put the device anywhere on the earth, orient it any way that you want, and conduct the experiment at any time of the year. In some orientation of the device, the veclocity of the aether will be in the $x$ direction, and in this direction, you would expect the time of flight difference between the legs to be maximized.

  • $\begingroup$ but could there not also be a velocity in the y-direction at this orientation which would affect the time difference? shouldn't this be included in the equation? $\endgroup$
    – physBa
    Sep 6, 2021 at 16:31
  • $\begingroup$ @physBa : the point is, if, in the diagram above, you have a velocity in the y direction, you simply rotate the apparatus until the new y component of the velocity is zero. $\endgroup$ Sep 6, 2021 at 16:38
  • $\begingroup$ Namely, the ether has some net velocity relative to the apparatus, and you're free to orient the apparatus however you wish. Pick the orientation that gives you the maximum signal. $\endgroup$ Sep 6, 2021 at 16:40
  • 1
    $\begingroup$ Oh yeah! There is always going to be a net velocity and the apparatus can be rotated so that the net velocity coincides with the x-axis. Ty $\endgroup$
    – physBa
    Sep 6, 2021 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.